Download Garrett 10-03-2011 1 We will later elaborate the ideas mentioned earlier: relations

Garrett 10-03-2011
1
We will later elaborate the ideas mentioned earlier: relations
of primes to zeros of zetas, reciprocity laws, p-adic and adelic
methods. Now... Commutative Algebra:
algebraic integer α ∈ Q: satisfies f (α) = 0, f ∈ Z[x] monic
Dedekind domains: unique factorization of ideals into prime ideals
integral extension of commutative rings O/o: every r ∈ O satisfies
f (r) = 0 for monic f ∈ o[x]
prime (ideal) P of O/o lying over prime ideal p of o, and residue
field extension O/P over o/p. Galois theory!
Helpful auxiliary ideas: localization S −1 o of a ring o to force
invertibility of elements of S, and v-adic completions ov , kv of o
and fraction field k, to squash field extensions of k.
Garrett 10-03-2011
An algebraic integer α ∈ Q satisfies f (α) = 0, for f ∈ Z[x] monic.
Also say α is integral over Z, or simply integral.
In a finite algebraic field extension k of Q, the ring (why!?!?)
o = ok of algebraic integers in k is
o = {α ∈ k : α is integral over Z}
√
Example: Inside quadratic field extensions k = Q( D) of Q,
√
with D a square-free integer. Reasonably-enough, α = a + b D
with a, b ∈ Z is integral, satisfying
α2 − 2aα + (a2 − b2 D) = 0
√
For D = 1 mod 4, there are more algebraic integers in Q( D) ...
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Let tr and N be Galois trace and norm k → Q. In terms of these,
we know the minimal polynomial for α is x2 − trα · x + N α. Thus,
in a quadratic extension, α is an algebraic√integer if and only both
trace and norm are in Z. Write α = a + b D with a, b ∈ Q.
The integrality condition is that 2a ∈ Z and a2 − b2 D ∈ Z. Try to
solve for rational integrality conditions on a, b.
From the first condition, at worst a ∈ 21 Z. With a = a0 /2 and
b = b0 /2, the second condition becomes a02 − b02 D ∈ 4Z.
Since the only squares mod 4 are 0, 1, for D = 2, 3 mod 4 actually
a0 , b0 ∈ 2Z, so a, b ∈ Z.
But for D = 1 mod 4, the condition is met for a0 = b0 mod 2!!!
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√
That is, the ring o of algebraic integers in k = Q( D) for squarefree integer D is
 √
(for D = 2, 3 mod 4)
 Z[ D]
o =
 1+√D
Z[ 2 ] (for D = 1 mod 4)
Indeed, we already knew
an example of the sense of this: the cube
√
−1+ −3
root of unity ω =
satisfies ω 2 + ω + 1 = 0.
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Caution: We will see that ignoring these ’extra’ algebraic
integers would be a fatal mistake: the resulting rings are very bad,
not Dedekind rings, exactly because they are not integrally closed,
that is, they omit elements of their fraction fields integral over Z.
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Example: Cyclotomic fields k = Q(ω), where ω is a primitive
nth root of unity. Since cyclotomic polynomials Φn are monic with
integer coefficients, certainly ω is an algebraic integer.
So the ring o of algebraic integers in k = Q(ω) contains Z[ω].
In fact, o = Z[ω], but this is not so easy to prove for n ≥ 5.
The sane proof uses ideas about localization, completion,
discriminant, different [sic], and ramification.
It is a fool’s errand to try to prove o = Z[ω] by writing out the
minimal polynomial of a + bω + cω 2 + . . . and examining the
integrality conditions.
Garrett 10-03-2011
Example:
Adjoining roots, for example, prime p-order
roots
√
√
p
p
k = Q( D) of square-free integers D. Certainly D is an
algebraic
integer, so the ring o of algebraic integers contains
√
p
Z[ D].
√
For D 6= 1 mod p2 , in fact, o = Z[ p D].
For D = 1 mod p√2 , in parallel with the square-root story, o is of
index p above Z[ 3 D], also containing
1+
√
p
D + ... +
p
√
p
D
p−1
√
3
For example, the ring o of integers in Q( 10) is
o = Z + Z·
√
3
10 + Z ·
1+
√
3
10 +
3
√
3
10
2
As with cyclotomic fields, it is unwise to try prove this directly.
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Why are these rings? Why are sums and products of algebraic
integers again integral?
This issue is similar to the issue of proving that sums and
products of algebraic numbers α, β (over Q, for example) are again
algebraic. Specifically, do not try to explicitly find a polynomial
P with rational coefficients and P (α + β) = 0, in terms of the
minimal polynomials of α, β.
The methodological point in the latter is first that it is not
required to explicitly determine the minimal polynomial of α + β.
Second, about algebraic extensions, to avoid computation,
recharacterization of the notion of being algebraic over... is
needed: an element α of a field extension K/k is algebraic over
k if k[α], the ring of values of polynomials on α, is a finitedimensional k-vectorspace.
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Recharacterization of integrality:
Let K/k be a field extension, o a ring in k with field of fractions
k.
We already know that α ∈ K is integral over o if f (α) = 0 for
monic f in o[x].
Claim: Integrality of α over o is equivalent to the condition that
there is a non-zero, finitely-generated o-module M inside K such
that αM ⊂ M .
Proof: On one hand, for α integral, with n = [k(α) : k], the omodule generated by 1, α, α2 , . . . , αn−1 is finitely-generated and is
stabilized by α...
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On the other hand, suppose αM ⊂ M , where M has o-generators
P
m1 , . . . , mn . Then there are cij ∈ o such that αmi = j cij mj ,
giving a system of n linear equations inside the field K:

 αm1 = c11 m1 + c12 m2 + . . . + c1n mn
...

αmn = cn1 m1 + cn2 m2 + . . . + cnn mn
or

c1n mn
 0 = (c11 − α)m1 + c12 m2 + . . . +
...

0 =
cn1 m1
+ c12 m2 + . . . + (cnn − α)mn
Existence of a non-zero solution m1 , . . . , mn implies vanishing of
determinant of

(c11 − α)
c12 . . .
c1n




...




cn1
c12
...
(cnn − α)
giving a monic equation satisfied by α !!!
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Corollary: In an algebraic field extension K/k, where k is the field
of fractions of a ring o, the set O of elements of K integral over o
is a ring.
Proof: Let α, β ∈ O, stabilizing non-zero, finitely-generated omodules M = hm1 , . . . , mµ i and N = hn1 , . . . , nν i. Then the omodule M · N generated by all products mi nj is non-zero, finitelygenerated, and is stabilized by α + β and by α · β (!)
///
Corollary: In the field extension Q/Q, the collection of all
algebraic integers really is a ring.
For a ring o inside a field K, the ring O of all elements of K
integral over o is the integral closure of o in K.
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Somewhat in parallel to development of the basics of algebraic
field theory, some unexciting things need to be checked. First,
from the monic-polynomial definition,
• For α ∈ K, an algebraic field extension of the field of fractions k
of o, for some 0 6= c ∈ o the multiple c · α is integral over o.
• For O integral over o, for any ring hom f sending O somewhere,
f (O) is integral over f (o).
Using the recharacterization:
• For O integral over o, if O is finitely-generated as an o-algebra,
then it is finitely-generated as an o-module.
• Transitivity: For rings A ⊂ B ⊂ C, if B is integral over A and C
is integral over B, then C is integral over A.
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Let’s prove the less-intuitive facts that need the
recharacterization:
For O finitely-generated as an o-algebra, use induction on the
number of algebra generators. This reduces to the step where
O = o[α], and α is integral over o. Ah! But proving that o[α] is
a finitely-generated o-module in this induction step is exactly the
recharacterization of integrality! Ha.
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Use the previous to prove the more interesting-sounding
transitivity of integrality. In A ⊂ B ⊂ C, any z ∈ C satisfies an
integral equation z n +bn−1 z n−1 +. . .+b1 z+bo = 0 with bi ∈ B. The
ring B 0 = A[bn−1 , . . . , bo ] is a finitely-generated A-algebra, so by
the previous it is a finitely-generated A-module. Since z satisfies
that monic, B 0 [z] is also a finitely-generated A-module. And since
z satisfies that monic, multiplication by z stabilizes B 0 [z]. The
latter is finitely-generated over A, so z is integral over A.
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Caution: Returning to the point that it would be a fatal mistake
to ignore the notion of integrality, for example, by discarding
algebraic numbers that are integral over Z, but meet naive
expectations:
Claim: UFD’s o are integrally closed (in their fraction fields k).
Proof: Let a/b be integral over o, satisfying
(a/b)n + cn−1 (a/b)n−1 + . . . + co = 0
with ci ∈ o. Multiplying out,
an + cn−1 an−1 b + . . . + bn co = 0
If a prime π in o divides b, then it divides a, by unique
factorization. Thus, taking a/b in lowest terms shows that b is a
unit.
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